Water Resources HubFE section 1 of 16 · free theory
Mathematics
The mathematics section is the engine room of the FE Civil exam: roughly one question in ten draws on it directly, and every other section leans on it quietly. Calculus, matrices, vectors, complex numbers, and a working feel for numerical methods will carry you through.
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Analytic geometry
Lines and conic sections show up as quick standalone questions and as the geometry behind calculus problems. Know the standard forms cold:
y = mx + b y − y1 = m(x − x1)
| m | slope of the line |
| b | y-intercept |
d = √((x2 − x1)² + (y2 − y1)²)
| d | distance between two points in the plane |
(x − h)² + (y − k)² = r²
| (h, k) | centre of the circle |
| r | radius |
y = ax² + bx + c, vertex at x = −b/(2a)
x²/a² + y²/b² = 1 (ellipse) x²/a² − y²/b² = 1 (hyperbola)
Read a conic from its equation: two squared terms with the same sign and equal coefficients → circle; same sign, unequal coefficients → ellipse; opposite signs → hyperbola.
Differential calculus
The exam tests differentiation as a tool: rates of change, slopes, and above all optimisation. If you can differentiate reliably and set the derivative to zero, half the calculus questions are already won:
d/dx xn = nxn−1 d/dx ex = ex d/dx ln x = 1/x
d/dx sin x = cos x d/dx cos x = −sin x d/dx tan x = sec²x
(uv)′ = u′v + uv′ (u/v)′ = (u′v − uv′)/v²
d/dx f(g(x)) = f′(g(x)) · g′(x) (chain rule)
f′(x) = 0 at a local max or min; f′′ > 0 → minimum, f′′ < 0 → maximum
Calculus trig derivatives assume x is in radians. If your calculator is in degree mode, every trig derivative answer will be wrong.
Integral calculus
Integration questions are usually definite integrals of polynomials, exponentials, or simple trig — evaluate the antiderivative at both limits and subtract:
∫xn dx = xn+1/(n+1) + C (n ≠ −1)
∫ex dx = ex + C ∫(1/x) dx = ln|x| + C
∫ab f(x) dx = F(b) − F(a)
∫u dv = uv − ∫v du (integration by parts)
f̄ = 1/(b − a) ∫ab f(x) dx (average value)
First-order differential equations
The FE keeps to two solvable types: separable equations and linear equations handled with an integrating factor. Spot the type first, then reach for the matching move:
dy/dx = f(x)·g(y) ⇒ ∫dy/g(y) = ∫f(x) dx (separable)
dy/dx + P(x)y = Q(x) (linear)
μ = e∫P(x) dx then d/dx(yμ) = Qμ
The integrating factor turns the left side into an exact derivative, d/dx(yμ). Multiply through, integrate both sides, then use the initial condition to pin down the constant C.
Matrices and systems of linear equations
Small systems (2×2, 3×3) are solved by determinants, elimination, or matrix inversion. Cramer's rule is the exam favourite for 2×2 because it is mechanical:
det a bc d = ad − bc
x = Dx/D, y = Dy/D (Cramer's rule)
| D | determinant of the coefficient matrix |
| Dx, Dy | determinant with the constant column swapped into column 1 (for x) or column 2 (for y) |
a bc d−1 = 1/(ad − bc) d −b−c a
Cramer's rule only works when D ≠ 0. If D = 0 the system is singular — no unique solution — and the exam will expect you to say so, not divide by zero.
Vector operations
The dot product measures alignment, the cross product measures area and gives a perpendicular direction. Keep straight which is a scalar and which is a vector:
A · B = a1b1 + a2b2 + a3b3 = |A||B| cos θ
|A × B| = |A||B| sin θ (direction by the right-hand rule)
A × B = −(B × A) A × A = 0
unit vector: uA = A/|A|, |A| = √(a1² + a2² + a3²)
Complex numbers
Complex arithmetic is pure bookkeeping once you accept i² = −1. Most FE questions are simple products, quotients via the conjugate, or modulus calculations:
(a + bi)(c + di) = (ac − bd) + (ad + bc)i
|z| = √(a² + b²), z̄ = a − bi, 1/z = z̄/|z|²
z = r(cos θ + i sin θ) = reiθ (polar form)
eiθ = cos θ + i sin θ (Euler's formula)
Numerical methods
When an equation cannot be solved in closed form, the exam expects you to know the standard numerical moves — root finding, quadrature, and finite differences — rather than to grind through dozens of iterations by hand:
Bisection: f(a)·f(b) < 0 ⇒ a root lies in (a, b); halve the interval each step
Newton: xn+1 = xn − f(xn)/f′(xn)
Trapezoidal: ∫ab f ≈ (h/2)(f0 + 2Σfi + fn)
Simpson's: ∫ab f ≈ (h/3)(f0 + 4Σfodd + 2Σfeven + fn), n even
Forward: (f(x+h) − f(x))/h Central: (f(x+h) − f(x−h))/(2h)
Central differences are markedly more accurate than forward differences for the same step size. Try all four methods yourself on the free numerical-methods calculator.
Worked example Definite integral of a polynomial
Given: Evaluate ∫03 (2x² + 3x − 1) dx.
Solution:
- Antiderivative term by term: 2x³/3 + 3x²/2 − x.
- At x = 3: 2(27)/3 = 18; 3(9)/2 = 13.5; −3. Sum: 18 + 13.5 − 3 = 28.5.
- At x = 0: 0. Subtract the lower limit: 28.5 − 0 = 28.5.
Answer: 28.5.
Worked example 2×2 system by Cramer's rule
Given:
- 2x + 3y = 8
- x − y = −1
Solution:
- Coefficient determinant: D = (2)(−1) − (3)(1) = −2 − 3 = −5 (non-zero, so a unique solution exists).
- Dx: replace column 1 with the constants: (8)(−1) − (3)(−1) = −8 + 3 = −5. So x = (−5)/(−5) = 1.
- Dy: replace column 2: (2)(−1) − (8)(1) = −2 − 8 = −10. So y = (−10)/(−5) = 2.
- Check: 2(1) + 3(2) = 8 ✓; 1 − 2 = −1 ✓.
Answer: x = 1, y = 2.
Worked example First-order linear ODE
Given: dy/dx + 2y = 6, with y(0) = 1. Find y(1).
Solution:
- This is linear with P(x) = 2, so the integrating factor is μ = e∫2 dx = e2x.
- Multiply through: d/dx(y·e2x) = 6e2x. Integrate: y·e2x = 3e2x + C, so y = 3 + Ce−2x.
- Apply y(0) = 1: 1 = 3 + C → C = −2. The particular solution is y = 3 − 2e−2x.
- At x = 1: y(1) = 3 − 2e−2 = 3 − 2(0.1353) = 3 − 0.2707 = 2.7293.
Answer: y(1) ≈ 2.73.